MHT CET · Physics · Waves and Sound
The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe of length ' \(L\) '. The length of the open pipe will be
- A \(\frac{L}{2}\)
- B \(\mathrm{L}\)
- C \(\mathrm{2L}\)
- D \(\mathrm{4L}\)
Answer & Solution
Correct Answer
(C) \(\mathrm{2L}\)
Step-by-step Solution
Detailed explanation
The length of closed pipe is denoted using \(\mathrm{L}\). Let \(l\) be the length of open pipe and \(\mathrm{v}\) be the velocity.
Frequency of second overtone of an open organ pipe is \(\mathrm{n}_{\mathrm{o}}=\frac{3 \mathrm{v}}{2 l}\)
Frequency of first overtone of a closed pipe is \(\mathrm{n}_{\mathrm{c}}=\frac{3 \mathrm{v}}{4 \mathrm{~L}}\)
Given: \(\mathrm{n}_0=\mathrm{n}_{\mathrm{c}}\)
\(\frac{3 \mathrm{v}}{2 l}=\frac{3 \mathrm{v}}{4 \mathrm{~L}}\)
\(\mathrm{L}=\frac{l}{2}\)
\(\therefore \quad l=2 \mathrm{~L}\)
Frequency of second overtone of an open organ pipe is \(\mathrm{n}_{\mathrm{o}}=\frac{3 \mathrm{v}}{2 l}\)
Frequency of first overtone of a closed pipe is \(\mathrm{n}_{\mathrm{c}}=\frac{3 \mathrm{v}}{4 \mathrm{~L}}\)
Given: \(\mathrm{n}_0=\mathrm{n}_{\mathrm{c}}\)
\(\frac{3 \mathrm{v}}{2 l}=\frac{3 \mathrm{v}}{4 \mathrm{~L}}\)
\(\mathrm{L}=\frac{l}{2}\)
\(\therefore \quad l=2 \mathrm{~L}\)
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